Fourier transform of an impulse-train sampled signal

Question

I'm trying to calculate the Fourier transform of an impulse-train sampled signal in two differnt ways but I end up with different results.

Impulse-train sampling of a continous signal $x(t)$ with sampling period $T_{s}$ is defined as

$$x_{p}(t) = x(t)p(t) \textrm{ where } p(t) = \sum_{n=-\infty}^{\infty} \delta(t - nT_{s})$$ Now consider the input signal $ x(t) = e^{-t}u(t)$ where the Fourier transform is $X(\omega) = \frac{1}{1 + j \omega}$. According to common procedure I calculate the Fourier transform of $x_{p}(t)$, using convolution in the frequency domain, as: $$X_{p}(\omega) = \frac{1}{T_{s}}\sum_{k=-\infty}^{\infty}X_{}(\omega- \frac{2\pi k}{T_{s}})$$ $$ \textrm{Inserting $X(\omega) $ I get } X_{p}(\omega)= \frac{1}{T_{s}}\sum_{k=-\infty}^{\infty}\frac{1}{1 + j (\omega - \frac{2\pi k}{T_{s}})} \tag{1}$$ The other way is to calculate $X_{p}$ directly from $x_{p}(t) = x(t)p(t) = \sum_{n=-\infty}^{\infty} x(nT_{s}) \delta(t - nT_{s})$. Now I get the result $$X_{p}(\omega) = \int_{-\infty}^{\infty} e^{-nT_{s}}\sum_{n=0}^{\infty} \delta(t - nT_{s}) e^{-j \omega t }dt = \sum_{n=0}^{\infty} e^{-nT_{s}(1+j \omega)} \tag{2}$$

I can't see that 1 and 2 are equal? Or are they? If not, what misstake did I do?

Answer 1

Both results are correct. Hence, the equality

$$\frac{1}{T}\sum_{k=-\infty}^{\infty}\frac{1}{1+ j \left(\omega - k\frac{2\pi }{T}\right)}=\sum_{n=0}^{\infty}e^{-nT(1+j\omega)}\tag{1}$$

holds true.

One way to prove this is to realize that the term on the left-hand side of Eq. $(1)$ is periodic with period $2\pi/T$. Consequently, we can express this term by its Fourier series

$$\frac{1}{T}\sum_{k=-\infty}^{\infty}\frac{1}{1+ j \left(\omega - k\frac{2\pi }{T}\right)}=\sum_{n=-\infty}^{\infty}c_ne^{jnT\omega}\tag{2}$$

It can be shown (try it!) that the Fourier coefficients $c_n$ are just (scaled) samples of the time domain function $x(t)$, so you obtain the right-hand side of Eq. $(1)$.

Eq. $(1)$ is an example of a more general result called Poisson's sum formula. The general form of Poisson's sum formula that specializes to Eq. $(1)$ is

$$\frac{1}{T}\sum_{k=-\infty}^{\infty}X\left(\omega-k\frac{2\pi}{T}\right)=\sum_{n=-\infty}^{\infty}x(nT)e^{-jnT\omega}\tag{3}$$

where $X(\omega)$ is the Fourier transform of $x(t)$.

Note that the right-hand side of Eq. $(3)$ is the discrete-time Fourier transform (DTFT) of the sequence $x(nT)$, which, according to Eq. $(3)$, equals the periodized spectrum $X(\omega)$ of the continuous-time function $x(t)$.

The dual form of Poisson's sum formula is

$$\sum_{k=-\infty}^{\infty}x(t-kT)=\frac{1}{T}\sum_{n=-\infty}^{\infty}X\left(n\frac{2\pi}{T}\right)e^{jn\frac{2\pi}{T}t}\tag{4}$$

i.e., the Fourier coefficients of a periodized time domain function $x(t)$ are samples of its Fourier transform $X(\omega)$.

Note that choosing $\omega=0$ in $(3)$, or, equivalently, $t=0$ in $(4)$ leads to the remarkable result

$$\sum_{k=-\infty}^{\infty}x(kT)=\frac{1}{T}\sum_{n=-\infty}^{\infty}X\left(n\frac{2\pi}{T}\right)\tag{5}$$